Inverse not always Unique for Non-Associative Operation

From ProofWiki
Jump to: navigation, search

Theorem

Let $\left({S, \circ}\right)$ be an algebraic structure.

Let $\circ$ be a non-associative operation.


Then for any $x \in S$, it is possible for $x$ to have more than one inverse element.


Proof

Proof by Counterexample:

Consider the algebraic structure $\left({S, \circ}\right)$ consisting of:

The set $S = \left\{{a, b, e}\right\}$
The binary operation $\circ$

whose Cayley table is given as follows:

$\begin{array}{c|cccc} \circ & e & a & b \\ \hline e & e & a & b \\ a & a & e & e \\ b & b & e & e \\ \end{array}$


By inspection, we see that $e$ is the identity element of $\left({S, \circ}\right)$.


We also note that:

\(\displaystyle \left({a \circ a}\right) \circ b\) \(=\) \(\displaystyle e \circ b\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle b\) $\quad$ $\quad$


\(\displaystyle a \circ \left({a \circ b}\right)\) \(=\) \(\displaystyle a \circ e\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle a\) $\quad$ $\quad$

and so $\circ$ is not associative.


Note further that:

\(\displaystyle a \circ b\) \(=\) \(\displaystyle e\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle b \circ a\) $\quad$ $\quad$

and also:

\(\displaystyle a \circ a\) \(=\) \(\displaystyle e\) $\quad$ $\quad$

So both $a$ and $b$ are inverses of $a$.

Hence the result.

$\blacksquare$


Also see


Sources